MHB Finding the Area bounded by the curve

[shamieh]

Find the area bounded by the curve $$x = 16 - y^4$$ and the y axis.

I need someone to check my work.

so I know this is a upside down parabola so I find the two x coordinates which are

$$16 - y^4 = 0$$
$$y^4 = 16$$
$$y^2 = +- \sqrt{4}$$
$$y = +- 2$$

so I know

$$\int^2_{-2} 16 - y^4 dy$$

Take antiderivative

$$16y - \frac{1}{5}y^5$$ | -2 to 2

so $$(2) = 32 - \frac{32}{5} = \frac{160}{5}$$

then $$(-2) = -32 - (\frac{-32}{5}) = -32 + \frac{32}{5} = \frac{-160}{5} + \frac{32}{5} = \frac{-128}{5}$$

SO finally
$$
[\frac{160}{5}] - [\frac{-128}{5}] = \frac{288}{5}$$

 

[alyafey22]

This is not a parabola. It is like a parabola that intersects the y-axis at $$y=\pm 2$$ so it is open to the left. I suggest you revise your calculations.

 

[MarkFL]

I would use the even-function rule to state:

$$A=2\int_0^2 16-y^4\,dy=\frac{2}{5}\left[80y-y^5 \right]_0^2=?$$

 

[shamieh]

Recalculated answer below if someone has a chance to check.

 

[shamieh]

recalculated and got $$\frac{256}{5}$$ . Is that correct?

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I would use the even-function rule to state:

$$A=2\int_0^2 16-y^4\,dy=\frac{2}{5}\left[80y-y^5 \right]_0^2=?$$

Yea this rule is so much easier!

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Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.

 

[shamieh]

Like how would I use this rule if i had $$5 - x^2 $$?

 

[MarkFL]

recalculated and got $$\frac{256}{5}$$ . Is that correct?

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Yea this rule is so much easier!

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Mark, know anywhere where I can find a good definition of the even function rule, so I can see how and when I can apply it etc. I googled it but couldn't find this one.

Yes, your result of $$A=\frac{256}{5}$$ is correct.

An even function is symmetric about the $y$-axis, i.e., $$f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.

Observe that:

$$\int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$

Now, in the first integral, if we replace $x$ with $-x$, we have:

$$\int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$

Bringing the negative in front of the differential out front and using $$f(-x)=f(x)$$, we have:

$$\int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,-dx+\int_0^a f(x)\,dx$$

Applying the FTOC, we obtain:

$$\int_{-a}^a f(x)\,dx=-\left(F(0)-F(a) \right)+F(a)-F(0)=2F(a)=2\int_0^a f(x)\,dx$$

 

[alyafey22]

Yes, your result of $$A=\frac{256}{5}$$ is correct.

An even function is symmetric about the $y$-axis, i.e., $$f(-x)=f(x)$$. If your limits of integration are also symmetric about the $y$-axis, then you may apply the even function rule.

Observe that:

$$\int_{-a}^a f(x)\,dx=\int_{-a}^0 f(x)\,dx+\int_0^a f(x)\,dx$$

Now, in the first integral, if we replace $x$ with $-x$, we have:

$$\int_{-a}^a f(x)\,dx=\int_{a}^0 f(-x)\,-dx+\int_0^a f(x)\,dx$$

Bringing the negative in front of the differential out front and using $$f(-x)=f(x)$$, we have:

$$\int_{-a}^a f(x)\,dx=-\int_{a}^0 f(x)\,dx+\int_0^a f(x)\,dx=\int_{0}^a f(x)\,dx+\int_0^a f(x)\,dx=2\int^a_0 f(x)\, dx $$